The gamma function is an analytic extension of the factorial function.

For real and positive $x$, with $x=int(x)+frac(x)$, where consider the function:

$$f(x)=\prod_\limits{i=1}^{int(x)} (frac(x)+i)$$

defined on $frac(x)\in[0,1)$.

We need to start from $i=1$ to avoid issues with $frac(x)=0$. This function mimics $x!$ in that $f(x)=x!$ for $x\in\mathbb{Z^+}$.

How can $\Gamma(x)-f(x)$ be explained?


1 Answer 1


I think that you can try something with the Bohr-Mollerup Theorem.

Theorem (Bohr-Mollerup): Let be $f:(0,+\infty)\to \Bbb{R}$ such that:

  1. $f(1)=1$ and $f(x)>0$, for every $x>0.$
  2. $f(x+1)=xf(x)$, for every $x>0.$
  3. $\lg{(f(x))}$ is a convex function.

Then $f\equiv \Gamma\big|_{(0,+\infty)}.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .